What are the maximal ideals $M$ of $R = \mathbb{F}_{p^n}[X,Y]/(XY - 1)$? What is the degree of $R/M$? Let $R = \mathbb{F}_{p^n}[X,Y]/(XY - 1)$. 


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*What are the maximal ideals $M$ of $R$?

*What does $R/M$ look like? What is it's degree over $\mathbb{F}_p$?
There is a theorem that states that for every $\mathbb{F}_p$-algebra $R$ of finite type and every maximal ideal $M \subset R$, $R/M$ is a finite field of characteristic $p$. 
Now I thought that the ring $R$ above is actually a field isomorphic to $\mathbb{F}_{p^n}(X)$. Since it's a field the only maximal ideal is the trivial one and hence $R/M = R / \{0\} \cong R$. But that is not finite which seems to contradict the theorem. Where did I go wrong? 
By the way I haven't yet found a proof of the mentioned theorem.
 A: Trying to write down an answer to an unanswered question. I'm also trying to clear to myself out some "obvious" things, as the proof, and write them down, so  please feel free to comment about anything in this post.
Putting $\,I:=\langle\,1-xy\,\rangle\,$ , define 
$$f:R\longrightarrow \mathbb F_{p^n}[x,x^{-1}]\,\,,\,\,f(g(x,y)+I):=g(x,x^{-1})$$
$\,f\,$ is well defined: $$g(x,y)+I=h(x,y)+I\Longrightarrow g(x,y)-h(x,y)=t(x,y)(xy-1)\Longrightarrow$$
$$ g(x,x^{-1})-h(x,x^{-1})=t(x,x^{-1})(xx^{-1}-1)=0\Longrightarrow g(x,x^{-1})=h(x,x^{-1})$$ 
$\,f\,$ is clearly a homomorphism of rings, and $\,g(x,x^{-1})=0\stackrel{(**)}\Longleftrightarrow g(x,y)\in I\,$ , from which we get that $\,f\,$ is bijective and thus an isomorphism.
Proof of $\,(**)\,$ above: let 
$$g(x,y)=\sum_{k=0}^ma_k(x)y^k\in \mathbb K[x][y]\cong \mathbb K[x,y]\,\,,\,\mathbb K=\,\text{any field}$$and write 
$$a_k(x)=\sum_{i=0}^{n_k}a_{ki}x^i\,\,,\,\,n_k\in\mathbb N\,\,,\,a_{ki}\in\mathbb K$$
Suppose $\,g(x,x^{-1})\equiv 0\,\,\,\text{in}\,\,\,\mathbb K[x,x^{-1}]$ . This means
$$g(x,x^{-1})=a_{0,0}+...+a_{0,n_0}x^{n_0}+x^{-1}\left(a_{1,0}+...+a_{1,n_1}x^{n_1}\right)+\ldots +x^{-m}\left(a_{n,0}+\ldots +a_{n,n_m}x^{n_m}\right)$$
The above expression being identically zero means that for any $\,k\in\mathbb N\,$ , we have
$$I\;\;\;\text{coefficient of}\,\,\,x^{-k}=a_{k,0}+a_{k+1,1} +\ldots +a_{j,j-k}+...=0$$ 
$$II\;\;\;\!\!\text{coefficient of}\,\,\,x^k\,\,=a_{0,k}+a_{1,k+1} +\ldots +a_{k,k+j}+...=0$$ 
The above sums are for sure finite, as there's only a finite number of powers (negative and positive) of $\,x\,$ with non-zero coefficients (this follows directly from the definition of ring of Laurent polynomials, of course).
Thus, going back to $\,g(x,y)\,$ and taking the sum of monomials with the above coefficients, we get 
$$a_{m,0}x^m+a_{m+1,1}x^{m+1}y+\ldots +a_{m+k,k}x^{m+k}y^k=x^m\left(a_{m,0}+a_{m+1,1}xy+...+a_{m+k,k}x^ky^k\right)=$$
$$\stackrel{by\,\,I}=x^m\left(-a_{m+1,1}-a_{m+2,2}-\ldots -a_{m+k,k}+a_{m+1,1}xy+\ldots a_{m+k,k}x^ky^k\right)=$$
$$=-x^m\left[a_{m+1,1}(1-xy)+a_{m+2,2}(1-x^2y^2)+...+a_{m+k,k}(1-x^ky^x)\right]\in I=\langle \,1-xy\,\rangle$$
since $\,1-x^ty^t=(1-xy)\left(1+xy+...+x^{t-1}y^{t-1}\right)\,$ . (The same argument works for the sum of monomials belonging to $\,II\,$ above)
